Hello, Mr_Green!
A ferris wheel with a diameter of 24 m rotates at a rate of 3 minutes/revolution.
Riders board the Ferris wheel 2 m above the ground.
A couple boards the Ferris wheel and the ride states.
(a) How far above the ground is the couple 30 seconds after the ride starts?
(b) How many seconds will it be (after the ride starts) before the couple
is 17 m above the ground for the second time?
So first, for (a) I went about developing the equation and i got:
. . h .= .12·cos(pi/90(t90)) + 14 .I don't follow this
so plug in 30 for t and get 8 meters above the ground, correct? .but this is correct!
and then i need some help with (b). Code:

* * *
*  *
*  *
*  *

*  *
* C *14 *
* θ\ 12 *
y \
*  \ *
* Q +* P
*  * :
* * * :h
2 :
     +    +     
O
C is the center of the ferris wheel, 14 feet above the ground: CO = 14.
The riders are at P, where CP makes angle θ with CO.
We see that: .CQ .= .12·cosθ
. . and that: .h .= .14  y .= .14  12·cosθ
The wheel turns 360° in 3 minutes: .120°/min.
. . In 30 seconds (½ minute), it has turned 60°.
(a) Therefore: .h .= .14  12·cos(60°) .= .14  12(½) .= .8 meters
When is h = 17? . 14  12·cosθ .= .17 . → . cosθ = 0.25
Then: .θ .≈ .104.5°, 255.5°, 464.5°, ...
Since this is the second time the couple is 17 m above the ground,
. . the angle is: .θ = 255.5°
At 120°/min, it takes: 255.5 ÷ 120 .= .2.12916666... minutes
(b) Therefore, it takes about 127.8 seconds.